Calculating gain, input and output resistances of a current amplifier circuit

This is (for the time being) more a comment than an answer.

The first KCL equation should be:

$$(V_x - U_{in})g_{be} + (V_x - V_y)g_{ce} + V_xG_e = (U_{in} - V_x)g$$

which simplifies to:

$$V_x(g_{be} + g_{ce} + G_e + g) -V_yg_{ce} = U_{in}(g_{be} + g) $$

For input resistance, find the current through \$r_{be}\$ which will be in terms of \$U_{in}\$.

The input resistance is then easily found to be:

$$r_{in} = \dfrac{U_{in}}{i_{r_{be}}}$$

First question: No, the circuit isn't perfectly symmetrical. The current mirror performs a differential- to single-ended conversion. If you wanted a perfectly symmetrical circuit, you would make M3 and M4 current sources, and then use some sort of common-mode feedback to set the appropriate current (so that \$V_{O1}\$ and \$V_{O2}\$ stay in a usable range).

The way it is now, \$A'_{v,dm}=\frac{g_m}{2g_{md}}\$ (where \$g_{md}\$ is the transconductance of the diode-connected device M3), while \$A''_{v,dm}=g_mr_{out}\$. Remember that the impedance seen looking into M3 is simply \$\frac{1}{g_{md}}\$. These two gains are obviously very different: \$A'_{v,dm}\ll A''_{v,dm}\$. The gain \$A'_{v,dm}\$ is so small that it's pretty useless, so the signal \$v_{o1}\$ is usually ignored, and a single, single-ended output on \$v_{o2}\$ is used.

Second question: I already said it, but \$A'_{v,dm}\$ is generally so small that \$v_{o1}\$ is ignored and isn't fed to the next gain stage (or output, or whatever follows). This means that the gain is simply \$A_{v,dm}=A''_{v,dm}\$. If you really want to, though, you can take both \$v_{o1}\$ and \$v_{o2}\$ as outputs to get \$A_{v,dm}=A'_{v,dm}+A''_{v,dm}\$.